3.30 \(\int \sec ^3(c+d x) (a+a \sec (c+d x))^4 \, dx\)

Optimal. Leaf size=136 \[ \frac {4 a^4 \tan ^5(c+d x)}{5 d}+\frac {4 a^4 \tan ^3(c+d x)}{d}+\frac {8 a^4 \tan (c+d x)}{d}+\frac {49 a^4 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^4 \tan (c+d x) \sec ^5(c+d x)}{6 d}+\frac {41 a^4 \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac {49 a^4 \tan (c+d x) \sec (c+d x)}{16 d} \]

[Out]

49/16*a^4*arctanh(sin(d*x+c))/d+8*a^4*tan(d*x+c)/d+49/16*a^4*sec(d*x+c)*tan(d*x+c)/d+41/24*a^4*sec(d*x+c)^3*ta
n(d*x+c)/d+1/6*a^4*sec(d*x+c)^5*tan(d*x+c)/d+4*a^4*tan(d*x+c)^3/d+4/5*a^4*tan(d*x+c)^5/d

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Rubi [A]  time = 0.17, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3791, 3768, 3770, 3767} \[ \frac {4 a^4 \tan ^5(c+d x)}{5 d}+\frac {4 a^4 \tan ^3(c+d x)}{d}+\frac {8 a^4 \tan (c+d x)}{d}+\frac {49 a^4 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^4 \tan (c+d x) \sec ^5(c+d x)}{6 d}+\frac {41 a^4 \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac {49 a^4 \tan (c+d x) \sec (c+d x)}{16 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^4,x]

[Out]

(49*a^4*ArcTanh[Sin[c + d*x]])/(16*d) + (8*a^4*Tan[c + d*x])/d + (49*a^4*Sec[c + d*x]*Tan[c + d*x])/(16*d) + (
41*a^4*Sec[c + d*x]^3*Tan[c + d*x])/(24*d) + (a^4*Sec[c + d*x]^5*Tan[c + d*x])/(6*d) + (4*a^4*Tan[c + d*x]^3)/
d + (4*a^4*Tan[c + d*x]^5)/(5*d)

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3791

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rubi steps

\begin {align*} \int \sec ^3(c+d x) (a+a \sec (c+d x))^4 \, dx &=\int \left (a^4 \sec ^3(c+d x)+4 a^4 \sec ^4(c+d x)+6 a^4 \sec ^5(c+d x)+4 a^4 \sec ^6(c+d x)+a^4 \sec ^7(c+d x)\right ) \, dx\\ &=a^4 \int \sec ^3(c+d x) \, dx+a^4 \int \sec ^7(c+d x) \, dx+\left (4 a^4\right ) \int \sec ^4(c+d x) \, dx+\left (4 a^4\right ) \int \sec ^6(c+d x) \, dx+\left (6 a^4\right ) \int \sec ^5(c+d x) \, dx\\ &=\frac {a^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {3 a^4 \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{2} a^4 \int \sec (c+d x) \, dx+\frac {1}{6} \left (5 a^4\right ) \int \sec ^5(c+d x) \, dx+\frac {1}{2} \left (9 a^4\right ) \int \sec ^3(c+d x) \, dx-\frac {\left (4 a^4\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d}-\frac {\left (4 a^4\right ) \operatorname {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{d}\\ &=\frac {a^4 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {8 a^4 \tan (c+d x)}{d}+\frac {11 a^4 \sec (c+d x) \tan (c+d x)}{4 d}+\frac {41 a^4 \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {a^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {4 a^4 \tan ^3(c+d x)}{d}+\frac {4 a^4 \tan ^5(c+d x)}{5 d}+\frac {1}{8} \left (5 a^4\right ) \int \sec ^3(c+d x) \, dx+\frac {1}{4} \left (9 a^4\right ) \int \sec (c+d x) \, dx\\ &=\frac {11 a^4 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {8 a^4 \tan (c+d x)}{d}+\frac {49 a^4 \sec (c+d x) \tan (c+d x)}{16 d}+\frac {41 a^4 \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {a^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {4 a^4 \tan ^3(c+d x)}{d}+\frac {4 a^4 \tan ^5(c+d x)}{5 d}+\frac {1}{16} \left (5 a^4\right ) \int \sec (c+d x) \, dx\\ &=\frac {49 a^4 \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {8 a^4 \tan (c+d x)}{d}+\frac {49 a^4 \sec (c+d x) \tan (c+d x)}{16 d}+\frac {41 a^4 \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {a^4 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {4 a^4 \tan ^3(c+d x)}{d}+\frac {4 a^4 \tan ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.82, size = 211, normalized size = 1.55 \[ -\frac {a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) \sec ^6(c+d x) \left (23520 \cos ^6(c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-\sec (c) (3750 \sin (2 c+d x)+15360 \sin (c+2 d x)-1920 \sin (3 c+2 d x)+3845 \sin (2 c+3 d x)+3845 \sin (4 c+3 d x)+6912 \sin (3 c+4 d x)+735 \sin (4 c+5 d x)+735 \sin (6 c+5 d x)+1152 \sin (5 c+6 d x)-11520 \sin (c)+3750 \sin (d x))\right )}{122880 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + a*Sec[c + d*x])^4,x]

[Out]

-1/122880*(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*Sec[c + d*x]^6*(23520*Cos[c + d*x]^6*(Log[Cos[(c + d*x)
/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]*(-11520*Sin[c] + 3750*Sin[d*x] +
3750*Sin[2*c + d*x] + 15360*Sin[c + 2*d*x] - 1920*Sin[3*c + 2*d*x] + 3845*Sin[2*c + 3*d*x] + 3845*Sin[4*c + 3*
d*x] + 6912*Sin[3*c + 4*d*x] + 735*Sin[4*c + 5*d*x] + 735*Sin[6*c + 5*d*x] + 1152*Sin[5*c + 6*d*x])))/d

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fricas [A]  time = 1.32, size = 137, normalized size = 1.01 \[ \frac {735 \, a^{4} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 735 \, a^{4} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (1152 \, a^{4} \cos \left (d x + c\right )^{5} + 735 \, a^{4} \cos \left (d x + c\right )^{4} + 576 \, a^{4} \cos \left (d x + c\right )^{3} + 410 \, a^{4} \cos \left (d x + c\right )^{2} + 192 \, a^{4} \cos \left (d x + c\right ) + 40 \, a^{4}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/480*(735*a^4*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 735*a^4*cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*(1152*
a^4*cos(d*x + c)^5 + 735*a^4*cos(d*x + c)^4 + 576*a^4*cos(d*x + c)^3 + 410*a^4*cos(d*x + c)^2 + 192*a^4*cos(d*
x + c) + 40*a^4)*sin(d*x + c))/(d*cos(d*x + c)^6)

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giac [A]  time = 3.64, size = 154, normalized size = 1.13 \[ \frac {735 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 735 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (735 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 4165 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 9702 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 11802 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 7355 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3105 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{6}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/240*(735*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 735*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(735*a^4*ta
n(1/2*d*x + 1/2*c)^11 - 4165*a^4*tan(1/2*d*x + 1/2*c)^9 + 9702*a^4*tan(1/2*d*x + 1/2*c)^7 - 11802*a^4*tan(1/2*
d*x + 1/2*c)^5 + 7355*a^4*tan(1/2*d*x + 1/2*c)^3 - 3105*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)
^6)/d

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maple [A]  time = 1.19, size = 146, normalized size = 1.07 \[ \frac {49 a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {49 a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {24 a^{4} \tan \left (d x +c \right )}{5 d}+\frac {12 a^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{5 d}+\frac {41 a^{4} \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{24 d}+\frac {4 a^{4} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {a^{4} \left (\sec ^{5}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{6 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+a*sec(d*x+c))^4,x)

[Out]

49/16*a^4*sec(d*x+c)*tan(d*x+c)/d+49/16/d*a^4*ln(sec(d*x+c)+tan(d*x+c))+24/5*a^4*tan(d*x+c)/d+12/5/d*a^4*tan(d
*x+c)*sec(d*x+c)^2+41/24*a^4*sec(d*x+c)^3*tan(d*x+c)/d+4/5/d*a^4*tan(d*x+c)*sec(d*x+c)^4+1/6*a^4*sec(d*x+c)^5*
tan(d*x+c)/d

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maxima [B]  time = 0.71, size = 270, normalized size = 1.99 \[ \frac {128 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{4} + 640 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{4} - 5 \, a^{4} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 120 \, a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

1/480*(128*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^4 + 640*(tan(d*x + c)^3 + 3*tan(d*x + c)
)*a^4 - 5*a^4*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 33*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4
+ 3*sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) - 180*a^4*(2*(3*sin(d*x + c)^3
- 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1))
- 120*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d

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mupad [B]  time = 4.65, size = 199, normalized size = 1.46 \[ \frac {49\,a^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,d}-\frac {\frac {49\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}-\frac {833\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+\frac {1617\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{20}-\frac {1967\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}+\frac {1471\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}-\frac {207\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^4/cos(c + d*x)^3,x)

[Out]

(49*a^4*atanh(tan(c/2 + (d*x)/2)))/(8*d) - ((1471*a^4*tan(c/2 + (d*x)/2)^3)/24 - (1967*a^4*tan(c/2 + (d*x)/2)^
5)/20 + (1617*a^4*tan(c/2 + (d*x)/2)^7)/20 - (833*a^4*tan(c/2 + (d*x)/2)^9)/24 + (49*a^4*tan(c/2 + (d*x)/2)^11
)/8 - (207*a^4*tan(c/2 + (d*x)/2))/8)/(d*(15*tan(c/2 + (d*x)/2)^4 - 6*tan(c/2 + (d*x)/2)^2 - 20*tan(c/2 + (d*x
)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 - 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{4} \left (\int \sec ^{3}{\left (c + d x \right )}\, dx + \int 4 \sec ^{4}{\left (c + d x \right )}\, dx + \int 6 \sec ^{5}{\left (c + d x \right )}\, dx + \int 4 \sec ^{6}{\left (c + d x \right )}\, dx + \int \sec ^{7}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+a*sec(d*x+c))**4,x)

[Out]

a**4*(Integral(sec(c + d*x)**3, x) + Integral(4*sec(c + d*x)**4, x) + Integral(6*sec(c + d*x)**5, x) + Integra
l(4*sec(c + d*x)**6, x) + Integral(sec(c + d*x)**7, x))

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